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Chemistry Question on Structure of atom

If λ0\lambda_{0} and λ\lambda be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

A

2hm(λ0−λ)\sqrt{\frac{2h}{m}\left(\lambda_{0}-\lambda\right)}

B

2hcm(λ0−λ)\sqrt{\frac{2hc}{m}\left(\lambda_{0}-\lambda\right)}

C

2hcm(λ0−λλ λ0)\sqrt{\frac{2hc}{m}\left(\frac{\lambda_{0}-\lambda}{\lambda\,\lambda_{0}}\right)}

D

2hm(1λ0−1λ)\sqrt{\frac{2h}{m}\left(\frac{1}{\lambda_{0}}-\frac{1}{\lambda}\right)}

Answer

2hcm(λ0−λλ λ0)\sqrt{\frac{2hc}{m}\left(\frac{\lambda_{0}-\lambda}{\lambda\,\lambda_{0}}\right)}

Explanation

Solution

E=W+12 mv2E=W+\frac{1}{2} \,mv^{2}
⇒hcλ=hcλ0+12mv2\Rightarrow \frac{hc}{\lambda}=\frac{hc}{\lambda_{0}}+\frac{1}{2} mv^{2}
⇒v2=2hcm[1λ−1λ0]\Rightarrow v^{2}=\frac{2hc}{m} \left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]
⇒v=2hcm[1λ−1λ0]\Rightarrow v=\sqrt{\frac{2hc}{m}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]}
⇒v=2hcm[λ0−λλλ0]\Rightarrow v=\sqrt{\frac{2hc}{m}\left[\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right]}